On Reduction, 9 of 11

Reciprocal Calculus

     Define a function’s “protractional fluxion”, or “pluxion”, this way:

     Pf   =  pf/px   =   lim  (f(x) <-> f(y)) / (x <-> y)

                         y->x

                      =   lim  (f(x<+>H) <-> f(x)) /  H

                          H->∞

     I think Newton’s term “fluxion” is a fine one, and it deserves more use; though here it is an abbreviation for “differential fluxion”. Where the differential fluxion is the ratio of two infinitesimal differences, the protractional fluxion is the ratio of two infinite protractions. Since the algebraic laws of reduction mirror those of addition, we can derive the following pluxion laws in the usual ways:

     Px      =   1

     PK      =   ∞           for any constant K

     P( K f)   =   K P(f)      for any constant K

     P(f<+>g)  =   P(f) <+> P(g)

     P(f<->g)  =   P(f) <-> P(g)

     P(f*g)  =   P(f)*g  <+>  f*P(g)

     P(1/g)  =   - P(g)/g²

     P(f/g)  =   ( P(f)*g  <-> f*P(g) ) / g²

     P(x²)   =   x/2

     P(x³)   =   x²/3

     And in general;       P(x^N)   =   x^N-1/N

     

     So we can protractionate reductive polynomials. For instance:

           P( 7 x³ <+> 5x² <-> 3x <+> 11 )  =  (7/3)x² <+> (5/2)x <-> 3

     Pluxions have the chain rule:

           pz/px       =      pz/py    *  py/px

           P(f(g(x)))   =  ((Pf)(g(x))) * (Pg)(x)

    

     Recall the Complementarity rule;

           (a-b)(a<->b) = -ab;

     protraction times difference equals negative product.

     Apply this to the product of pluxion and fluxion:

     For y near to x, (pf/px)*(df/dx) approximately equals:

          {(f(x)<->f(y))/ (x<->y) }*{(f(x)-f(y))/ (x-y) }

     =    {(f(x)<->f(y))*(f(x)-f(y))} / {(x<->y)*(x-y)}

     =    { - f(x) * f(y) } / { - x * y }

     Which in the limit approaches

           (f(x))² / x²

     Therefore in general:

           (pf/px)*(df/dx)  =   (f(x))² / x²

           Pf * f’         =    f²/x²

     Therefore the Pluxion from Fluxion law:

           pf/px   =    (f(x))² / ( x² * df/dx )

           Pf      =     f² /(x^{2 .} f’)

     In particular:

     P(e^x)         =     e^x /x²

     P(e^-1/x)       =     e^-1/x

     P(ln(x))      =    (ln(x))² / x

     P(-1/ln(x))   =     1 / x

     We can now derive P(f+g):

           P(f+g)     =    (f+g)² / (x²(f+g)’)

                      =    (f+g)² / (x²(f’+g’))

                      =    (f+g)²( 1/(x²f’)  <+>  1/(x²g’) )

                      =    (f+g)²  ( Pf/f²  <+>  Pg/g² )

     Similarly;

           (f<+>g)’   =    (f<+>g)²  ( f’/f²  +  g’/g² )

     For instance;

           (x<+>x²)’  =    (x<+>x²)²  ( x’/x²  +  x²’/(x²)² ) 

                      =    x²(1<+>x)² ( 1/x²   +  2x/x⁴ ) 

                      =    (x²*x²/(x+1)²)( 1/x²   +  2/x³ ) 

                      =    (x⁴/(x+1)²)( (x + 2) / x³ ) 

                      =    x(x+2)/(x+1)²

           P(x+x²)    =    (x+x²)²  ( Px/x²  <+>  Px²/(x²)² ) 

                      =    x²(1+x)² ( 1/x²   <+>  (x/2)/x⁴ ) 

                      =    x²(1+x)² ( 1/x²   <+>  (1/2x³ ) 

                      =    (1+x)² ( 1  <+>  (1/2x) ) 

                      =    (1+x)² ( 1 / (1+2x) ) 

                      =    (1+x)² / (1+2x)  

     For longer sums we have:

     P(f+g+h)   =    (f+g+h)²  ( Pf/f²  <+>  Pg/g²  <+>  Ph/h²)

     (f<+>g<+>h)’    =    (f<+>g<+>h)²  ( f’/f²  +  g’/g²  +  h’/h²)

     And so on.

     One of the uses of the fluxion is in Newton’s Method for finding the roots of a function. To solve f(x) = 0, start with an estimate x₁, then run this iteration:

           X_N+1   =   X_N – f(X_N)/f’(X_N)

     But   Pf   =   f² /(x^{2 .} f’)

     so   f/f’ =    x² Pf / f

     and therefore

           X_N+1   =   X_N –  X_N² P(X_N)/f(X_N)

          X_N+1   =   X_N (  1  –  X_N*Pf(X_N)/f(X_N) )

     This is the pluxion version of Newton’s Method.

     Another use of the fluxion is finding extrema. But since Pf*f’ = f²/x², f’>0 only if Pf>0; f’=0 only if Pf=∞; and f’<0 only if Pf<0. So you may find critical points by solving  Pf(x)= ∞;  then examine Pf near the critical point to tell if it is an extremum or not. If Pf goes from – through ∞ to +, then the critical point is a minimum; if Pf goes from + through ∞ to – then the critical point is a maximum. This is the First Pluxion Test.

     Inverse to the Pluxion is the Indefinite Reciprocal Integral, also called the Anti-pluxion:

           R( f(x) px )  =  F(x) <+> C   if and only if   P(F(x)) = f(x)

           C is the constant of reciprocal integration.

     There’s also a Definite Reciprocal Integral:

           R_a^b(f(x) px)  = 

lim  ( f(a)<+>f(a<+>H)<+>f(a<+>H/2)<+>f(a<+>H/3)<+> … <+>f(b) )* H )

H->∞

     These two are united by the Fundamental Theorem of Reciprocal Calculus:

           If    F = R_a^x(f(y) py)    then     pF/px = f

           If   pF/px  =  f    then    R_a^b(f(x) px) =  F(b) y F(a)

     Here is a short table of reciprocal integrals:

           R( 1 px )      =      x

           R( ∞ px )      =      C  ,  a constant.

           R(K*f px )      =     K * R(f px)    ,   for any constant K.

           R((f <+> g)px )   =        R(f px) <+> R(g px)

           R((f <-> g)px )   =        R(f px) <-> R(g px)

           R( x^N px )      =      (N+1) x^N+1

           R( (1/x) px )   =     -1 / ln(x)   =  log_x(1/e)

           R( e^-1/x  px )   =       e^-1/x 

     Here’s the Chain Rule:

    

           R( f(g(x)) Pg(x) px )   =   (Rf)(g(x))

     Here’s Reductive Integration by Parts:

           R( Pf * g * px )    =   f*g  <->  R( f * Pg * px )

          

           R ( g * pf )        =   f*g  <->  R( f * pg )